All categories

3 years ago

Which is hotter, Canopus or Vega? How much brighter?

1 answer:

4 0

Temperature determines the hotness or coldness. The surface temperature of Canopus is 7350 K. The surface temperature of Vega is 9602 K. Because the surface temperature of Vega is greater than Canopus, Vega is hotter.

Brightness is determined by the star's magnitude. The apparent magnitude of Canopus is (-0.72) and that of Vega is (0.03). we know that, lesser the magnitude, brighter is the star. Hence, Canopus is brighter.

Vega is hotter but Canopus is brighter. This is because, brightness depends on the star's distance and size of the star as well other than the factor temperature.

You might be interested in

hooke's law is described mathematically using the formula fsp=-kx. which statement is correct about spring force, fsp? A. It is

bogdanovich [222]

Answer:

A. It is always a positive force

Explanation:

Hooke's law describes the relation between an applied force and extension ability of an elastic material. The law states that provided the elastic limit, e, of a material is not exceeded, the force, F, applied is proportional to the extension, x, provided temperature is constant.

i.e F = - kx

where k is the constant of proportionality, and the minus sign implies that the force is a restoring force.

The applied force can either be compressing or stretching force.

6 0

3 years ago

.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in c

ohaa [14]

Answer:

a = 1.1 10⁵ m / s²

Explanation:

This is a momentum exercise, where we use the relationship between momentum and momentum

I = ∫ F dt = Δp

= p_f - p₀

as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction

F t = 2 (m v)

therefore the average force is

F = 2 m v / t

where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy

starting point. Done the ball is released with zero initial velocity

Em₀ = U = mgh

final point. Upon reaching the ground, just before the deformation begins

Em_f = K = ½ m v²

energy is conserved in this system

Em₀ = Em_f

m g h = ½ m v²

v = √ (2gh)

This is the velocity of the body when it reaches the ground, so the force remains

F = 2m √(2gh) /t

where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds

knowing this force let's use Newton's second law

F = m a

a = F / m

a = 2 √(2gh) / t

We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s

a = 2 √ (2 9.8 1.5) / 10⁻³

a = 1.1 10⁵ m / s²

6 0

3 years ago

Question 12 (1 point) Question 12 Unsaved

Bezzdna [24]

Slope is your answer

7 0

3 years ago

A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no apprecia

masha68 [24]

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

$V(t) = \epsilon. e^{-t.\frac{L}{R} }$

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

$0.8*19 = 19. e^{-4.\frac{22}{R} }$

$0.8 = e^{-\frac{88}{R} }$

$ln(0.8) = ln(e^{-\frac{88}{R} })$

$ln(0.8) = -\frac{88}{R}$

$R = -\frac{88}{ln(0.8)}$

R = 394.36

In this LR circuit, the resistance of the resistor is 394.36ohms.

7 0

3 years ago

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Sedbober [7]

Answer:

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

$\omega =\sqrt{\frac{k}{m}}$

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

$x=A\cos(\omega t)$

Finally, the equation of the motion of the system is:

$x=(0.088m)\cos(\omega t)$

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

7 0

3 years ago